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Part I: Increasing and Decreasing Functions and Signs of Tangent Line Slopes 2. Part II: Slopes of Tangent Lines and Local Extrema 3. Part III: Tangent lines above and below the curve and concavity. 4. Part IV: Points of inflection and concavity 5. Extensions: I. Position, Velocity, and Acceleration II. Linear Approximations to Functions Solutions Introduction to Tangent Lines using TI-nspire CAS Summary of Activity: Students will use the TI-Nspire to draw the tangent lines to a point on a particular graph, grab the point and move it around on the graph, and observe the changes in the tangent line. Students will be asked to observe the relationship of the tangent to the curve, the signs of the slopes of the tangent lines, slopes of tangents at extrema, and the position of the tangents at points of inflection and concavity. Assumptions of Previous Knowledge Students can solve and find derivatives algebraically and know that the definition of derivative is the slope of the tangent line. This activity is designed to be an introduction to curve sketching, finding extrema, identifying where a function increases and decreases, determining concavity, and finding points of inflection that bridges the two approaches. Students must also be able to tell on what intervals a function is increasing or decreasing and if a line is above or below the curve. They must also understand what minimums and maximums are on a graph. Students must also be familiar with the basic operations of the TI-Nspire, specifically graphing a function, drawing a tangent to the curve, measuring the slope and dragging the tangent around. Directions are given for these procedures for the first problem only. Extension to the Activity Two extensions are included. The first one asks students to explore the relationship between position, velocity, and acceleration. The second focuses on linear approximations and whether the estimates are underestimates or overestimates. Ohio State Standards Grade: 12 Strand: Patterns, Functions, and Algebra #7: Make mathematical arguments using the concept of limit #10 Use the concept of limits to find instantaneous rate of change for a point on a graph as a slope of tangent at a point. A.P. Objectives This activity emphasizes the Advanced Placement objectives listed below. These objectives are found in the AP Course Description handbook distributed to AP teachers. The slope of a curve at a point. Examples are emphasized, including points at which there are vertical tangents and points at which there are no tangents. Tangent line to a curve at a point and local linear approximation Relationship between the concavity of ƒ and the sign of ƒ Points of inflection as places where concavity changes Corresponding characteristics of graphs of ƒ and ƒ Relationship between the increasing and decreasing behavior of ƒ and the sign of f Introduction to Tangent Lines using TI-Nspire CAS Student Activity Sheet Objectives: Students will graph a function, draw the tangent line to a point, move the tangent along the function, and make observations regarding the relationship of the tangent to the curve, the signs of the slopes of the tangent lines, slopes of tangents at extrema, and the position of the tangents at points of inflection and concavity. To the student: You have been finding the derivative of a variety of functions using Calculus rules such as the power rule, product and quotient rules, and the chain rule. You have also found the slope of a tangent line at a point and used it to write the equation of the tangent line. This activity will help you explore the behavior of tangents and the concept of the derivative. Part One: Increasing and Decreasing Functions and the Signs of Slopes Using the TI n-spire, graph QUOTE  . Set the window to [-10,10] by [-10,10]. Choose the home icon, select 2:Graphs. Enter f1(x)= x2 Choose menu button, select 4:Window’!1:Window Settings. Set the x-min to -10, hit TAB. Set the x-max to 10 and hit TAB. Set the y-min to -10 and hit TAB. Set the y-max to 10, and hit enter. Make sure calculator is in radian mode. Now draw a point on the curve  QUOTE   Choose the menu icon, select 6:Points and Lines’!2:Point on Move the cursor to the curve until “point on” and a number appears. Hit enter and the point should appear and coordinates should darken. Hit the esc key. You can move the coordinates and drag them to an empty area of the graph, if you wish: Move the cursor to the coordinates until “text” appears. Hold the hand button down until the finger and thumb pinch together and drag it to an empty portion on the graph. Hit esc key to release. Now draw a tangent to the curve  QUOTE   at the point you drew in step 2. Choose the menu icon, select 6:Points and Lines’!7:Tangent Move the cursor to the curve until  point on appears. Hit enter and the tangent line should appear. Hit the esc key. Next calculate the slope of the tangent line you drew in step 3. Choose the menu icon, select 7:Measurement’!3:Slope Move the cursor to the tangent line until  line appears and hit enter twice. The number should darken. Hit the esc key. (You can move this number to an empty area of the graph by following the directions at the end of step 2.) You can add text to the screen using the menu icon. Select 1: Actions’!5 Text. Your Screen should now look like the screen at the right. Drag the point around the graph and the tangents should move as well. Move the point around and observe what happens to the tangents and the slope of the tangent line. You will use this process to fill out the tables in this activity. Move the cursor to the point until the word “point” appears. Hold down the hand button until the finger and thumb pinch together and the point is flashing. Use the navigation pad to move the point along the graph. If you wish, hit the esc key to release. Fill out the chart below. Move the point close to the coordinates listed in the table.. You may or may not be able to get the actual coordinate point, but get AS CLOSE AS YOU CAN.. Fill in positive or negative for the sign of the slope. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = -3x = -2x = -1x = -0.5x = 0.5x = 1x = 2x = 3 Now, on a new graph page, draw the graph of  QUOTE  . Keep the same window setting. Repeat the process for drawing a tangent at a point for the function  QUOTE  . Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Choose the home icon, select 2:Graphs. This gives you a new graph as well as a new page. Enter  QUOTE   Choose menu button, select 4:Window’!1:Window Settings. Set the x-min to -10, hit TAB. Set the x-max to 10 and hit TAB. Set the y-min to -10 and hit TAB. Set the y-max to 10, and hit enter. Follow the rest of steps 2-5. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = -3x = -2x = -1x = -0.5x = 0.5x = 1x = 2x = 3 NOTE: For each of the new tables that follow, begin the graph on a new graph page. Repeat the process for drawing a tangent at a point for the function y = x3, keeping the same window settings. Enter the results in the table below. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = -2x = -1.5x = -1x = -0.5x = 0.5x = 1x = 1.5x = 2 Repeat the process for drawing a tangent at a point for the function y = ex, keeping the same window settings. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = -4x = -2x = -1x = -0.5x = 0.5x = 1x = 1.5x = 2 Repeat the process for drawing a tangent at a point for the function QUOTE  , changing the window settings to [-1, 35] by [-5, 5]. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = 0.5x = 1x = 3x = 10x = 15x = 20x = 25x = 30 Repeat the process for drawing a tangent at a point for the function.  QUOTE   Use the window settings [-3, 6] by [-2, 2]. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = x3 increasing or decreasing near that point?Sign of the slope of the tangent linex = -2.7x = -2.5x = -1x = -0.5x = 0.5x = 1x = 2.5x = 3x = 4x = 5 Looking at all of the charts, what conclusion can be made regarding the sign of the slope of the tangent line: (a) When the function is increasing? (b) When the function is decreasing? Where do you think that the sign of the tangent switches from positive to negative? Without graphing, can you tell if a function is increasing or decreasing by looking a given point on the function? Consider y = 2x2 + 4x – 3. Take the derivative by hand at the points x = -2 and at x = 3. Is it increasing or decreasing at each of those points? Show all of your work. Part Two: Slopes of Tangents and Local Extrema NOTE: Local extrema are the minimums or maximums of a function on a given interval. Graph  QUOTE  , on a new graph screen using a window setting of [-10, 10] by [-10, 10]. Draw a point on the curve and draw a tangent to that point. Drag the tangent to the points where the graphs reach a minimum value. Fill in the chart below: Refer to steps 1-5 from Part One. Graph the point and drag it to the minimum value. You know you are at the minimum because a lower case m in a box appears to the lower right of the “hand.” Remember that if you cannot read some of the information because it is all bunched together, you can drag the coordinates and slopes to another area of the screen. Refer to Step 2 in Part One. Coordinates of MinimumValue of the slope of the tangent at the minimumThe tangent line is increasing, decreasing, horizontal, or vertical… Graph  QUOTE   on a new graph screen using, the same window settings. Draw a point on the curve and draw a tangent to that point. Drag the tangent to the points where the graphs reach a maximum value. Fill in the chart below: Graph the point and drag it to the maximum value. You know you are at the maximum because an upper case M in a box appears to the lower right of the “hand.” Coordinates of MaximumValue of the slope of the tangent at the minimumThe tangent line is increasing, decreasing, horizontal, or vertical… Graph  QUOTE   on a new graph screen using the window [-3, 6] by [-2, 2]. Draw a point on the curve and draw a tangent to that point. Drag the tangents to the points where the graphs reach a maximum or minimum value. Fill in the chart below: Min or MaxCoordinates of Min or MaxValue of the slope of the tangent at the minimumThe tangent line is increasing, decreasing, horizontal, or vertical… Now make a conclusion about your observations from #1-3. Fill in the blanks: At an extrema, the slope of the tangent is ______ and the tangent line is ___________. Now we want to examine the behavior of a function before and after an extrema point. Consider  QUOTE   . From part two, #1, the coordinates of the minimum were (0, 0). Fill in the table below, using information from part one and part two, as well as your graph: intervalf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-", 0)(0, ") Conclusion: Is the point (0, 0) a minimum or maximum? Justify your answer. Consider  QUOTE  . From part two, #2, the coordinates of the maximum were (0, 0). Fill in the table below, using information from Part One and Part Two, as well as your graph: intervalf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-", 0)(0, ") Conclusion: The point (0, 0) is a minimum or maximum? Justify your answer. The function  QUOTE  on the given interval had two minimums and one maximum. From part 2 #3, what were the coordinates of these three points? Minimums: Maximum: Fill in the table below, using information from part one and part two, as well as your graph: intervalsf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-3, -1.5708)(-1.5708, 1.5708)(1.5708, 4.7123)(4.7123, 6) Make a conclusion about determining if a point is a maximum or minimum based on your observations: A point is a maximum if the slope of the tangent to the point is __________ and the slope of the tangents switch from ______________ before the point to _________________after the point. A point is a minimum if the slope of the tangent to the point is __________ and the slope of the tangents switch from ______________ before the point to _________________after the point. Consider the other functions from part 1: y = x3, y = ex, and  QUOTE  . Look at the graphs and your tables from Part One. Do these functions have any extrema? If so, are they minimums or maximums? Explain your answer based on your conclusions from Part Two, #8. How can you find extrema without looking at the graph? Consider y = 2x2 + 4x – 3. Take the derivative by hand. Based on your answers from #8, find where an extrema occurs. Determine if the extrema is a maximum or minimum. Show all work and explain your answer. Part Three: Tangents Above or Below the Curve and Concavity Tangents can lie above or below the curve. All tangent lines that lie below the curve have y-values that are less than the corresponding y-values on the function. In other words, if f(x) are the y-values on the curve and T(x) are the y-values on the tangent curve, then T(x) d" f(x) for all values of x. All tangents lines that lie above the curve have y-values that are more than the corresponding y-values on the functions. In other words, T(x) e" f(x) for all values of x. Let s revisit the function  QUOTE  . As you move the point along the graph, are the tangents above the graph or below the graph? What is the general shape of the function, specifically looking at whether it opens upward or downward? A graph that opens upward is referred to as being concave up. A graph that opens downward is concave down. Is  QUOTE   concave up or concave down? Now look at  QUOTE  . As you move the point along the graph, are the tangents above the graph or below the graph? Is  QUOTE   concave up or down? Explain. Now look at y = ex. As you move the point along the graph, are the tangents above the graph or below the graph? Is y = ex concave up or down? Explain. Now look at  QUOTE  . As you move the point along the graph, are the tangents above the graph or below the graph? Is  QUOTE   concave up or down? Explain. Now make a conclusion: With respect to tangent lines, a function is concave up when: ____________________________________________________________________________. With respect to tangent lines, a function if concave down when: ____________________________________________________________________________. Part Four: Points of Inflection and Concavity Some functions won’t have a single concavity. Let’s look at some examples of those functions. As you move the point along the graph, the tangents lines may switch positions. The point at which the concavity switches is called the point of inflection. At this point, some of the points of the tangent line are above the curve and some of the points are below. Consider y = x3. Drag the tangents along the curve. Find at which point on y = x3 that tangents lines switch from above to below the curve or vice-versa? On what interval is y = x3 concave up? On what interval is y = x3 concave down? Now, consider  QUOTE   on [-3, 6] by [-2, 2]. On what interval(s) is  QUOTE   concave up and on what interval(s) is  QUOTE   concave down? What are the points of inflection? Extension One: Position, Acceleration, and Velocity The velocity of a particle is defined to be the rate of change in position with respect to time. Specifically, instantaneous velocity at any time is given by the equation  QUOTE   The acceleration of a particle is defined to be the rate of change in velocity with respect to time. Specifically, instantaneous acceleration at any time is given by the equation  QUOTE   A particle is said to be speeding up when its velocity and acceleration are in the same direction (both positive or both negative). A particle is said to be slowing down when its velocity and acceleration are in opposite directions (either positive or negative). A particle moving in the y-direction has a position given at any time t by the function:  QUOTE  . (Suggested viewing window for this problem is [-1, 12] x [-40, 50]) During the first 8 seconds of the particle’s motion: When is the object falling? When is the object rising? When does the object reach its maximum height? When does the object reach its minimum height? When is its velocity increasing? When is its velocity decreasing? When does the object reach maximum velocity? Extension Two: Linear Approximation and Tangent Lines Graph the equation y = x3 + 2x2 - 3x – 4 on the window [-10, 10] by [-10, -10]. Draw a point on the curve, draw a tangent line through that point, and find the equation of the tangent line, using the TI-Nspire. To draw a point on the curve and draw a tangent through that point, refer to Activity 1. For the equation of the tangent, choose the menu icon, select 1:Action’!6:Coordinates and Equations Move the cursor to the point and select enter and the line will be highlighted and the equation appears lightly. Hit enter a second time and the equation will darken. Hit the esc key. (You can drag this equation to an empty area of the graph by following the directions in Activity 1) We know the point (1, 2) is on the curve. The equation to the tangent line at the point x=1 is given by  QUOTE   We can use this tangent line to approximate values of the function close to the point x=1. For example, the estimate,  QUOTE   Notice that actual value,  QUOTE   Also notice that the estimate,  QUOTE  , whereas the actual value,  QUOTE  . In both cases, the estimates are less than the actual values. Looking at the tangent line at x = 1, we could have predicted this underestimation. Why? Now consider the point (-2, 2) on the curve. The equation of the tangent line at this point is  QUOTE   What is the linear approximation of  QUOTE   at the point x = -2.08? What is the actual value at this point? Similarly for the point x = -1.997: what is the linear approximation and the actual value? Are the two linear estimates over or under? What do we notice about the relationship between the tangent line and the curve at this point? What can you conclude about linear approximations using tangents and the concavity of the curve? Solutions Introduction to Tangent Lines using TI-nspire CAS Student Activity Sheet Objectives: Students will graph a function, draw the tangent line to a point, move the tangent along the function, and make observations regarding the relationship of the tangent to the curve, the signs of the slopes of the tangent lines, slopes of tangents at extrema, and the position of the tangents at points of inflection and concavity. To the student: You have been finding the derivative of a variety of functions using Calculus rules such as the power rule, product and quotient rules, and the chain rule. You have also found the slope of a tangent line at a point and used it to write the equation of the tangent line. This activity will help you explore the behavior of tangents and the concept of the derivative. Part One: Increasing and Decreasing Functions and the Signs of Slopes Using the TI-Nspire, graph QUOTE  . Set the window to [-10,10] by [-10,10]. Choose the home icon, select 2:Graphs Enter f1(x)= x2 Choose menu button, select 4:Window’!1:Window Settings. Set the x-min to -10, hit TAB. Set the x-max to 10 and hit TAB. Set the y-min to -10 and hit TAB. Set the y-max to 10, and hit enter. Now draw a point on the curve y = x2 Choose the menu icon, select 6:Points and Lines’!2:Point on Move the cursor to the curve until “point on” and a number appears. Hit enter and the point should appear and coordinates should darken. Hit the esc key. You can move the coordinates and drag them to an empty area of the graph, if you wish: Move the cursor to the coordinates until “text” appears. Hold the hand button down until the finger and thumb pinch together and drag it to an empty portion on the graph. Hit esc key to release. Now draw a tangent to the curve y = x2 at the point you drew in step 2. Choose the menu icon, select 6:Points and Lines’!7:Tangent Move the cursor to the curve until  point on appears. Hit enter and the tangent line should appear. Hit the esc key. Next calculate the slope of the tangent line you drew in step 3. Choose the menu icon, select 7:Measurement’!3:Slope Move the cursor to the tangent line until  line appears and hit enter twice. The number should darken. Hit the esc key. (You can move this number to an empty area of the graph by following the directions at the end of step 2.) You can add text to the screen using the menu icon. Select 1: Actions’!5 Text. Your Screen should now look like the screen at the right Drag the point around the graph and the tangents should move as well. Move the point around and observe what happens to the tangents and the slope of the tangent line. You will use this process to fill out the tables in this activity. Move the cursor to the point until the word “point” appears. Hold down the hand button until the finger and thumb pinch together and the point is flashing. Use the navigation pad to move the point along the graph. If you wish, hit the esc key to release. Fill out the chart below. Move the point as close to the coordinates listed in the table, AS CLOSE AS YOU CAN. You may or may not be able to get the actual coordinate point, but get as close as you can. Fill in positive or negative for the sign of the slope. Coordinate PointIs y = x2 /4 increasing or decreasing?Sign of the slope of the tangent linex = -3decreasingnegativex = -2decreasingnegativex = -1decreasingnegativex = -0.5decreasingnegativex = 0.5increasingpositivex = 1increasingpositivex = 2increasingpositivex = 3increasingpositive Now, on a new graph page, draw the graph of  QUOTE  . Keep the same window setting. Repeat the process for drawing a tangent at a point for the function  QUOTE  . Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Choose the home icon, select 2:Graphs. This gives you a new graph as well as a new page. Enter  QUOTE   Choose menu button, select 4:Window’!1:Window Settings. Set the x-min to -10, hit TAB. Set the x-max to 10 and hit TAB. Set the y-min to -10 and hit TAB. Set the y-max to 10, and hit enter. Follow the rest of steps 2-5. Coordinate PointIs y = -x2/4 increasing or decreasing?Sign of the slope of the tangent linex = -3increasingpositivex = -2increasingpositivex = -1increasingpositivex = -0.5increasingpositivex = 0.5decreasingnegativex = 1decreasingnegativex = 2decreasingnegativex = 3decreasingnegative NOTE: For each of the new tables that follow, begin the graph on a new graph page. Repeat the process for drawing a tangent at a point for the function y = x3, keeping the same window settings. Enter the results in the table below. Coordinate PointIs y = x3 increasing or decreasing?Sign of the slope of the tangent linex = -2increasingpositivex = -1.5increasingpositivex = -1increasingpositivex = -0.5increasingpositivex = 0.5increasingpositivex = 1increasingpositivex = 1.5increasingpositivex = 2increasingpositive Repeat the process for drawing a tangent at a point for the function y = ex, keeping the same window settings. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = ex increasing or decreasing?Sign of the slope of the tangent linex = -4increasingpositivex = -2increasingpositivex = -1increasingpositivex = -0.5increasingpositivex = 0.5increasingpositivex = 1increasingpositivex = 1.5increasingpositivex = 2increasingpositive Repeat the process for drawing a tangent at a point for the function QUOTE  , changing the window settings to [-1, 35] by [-5, 5]. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = ln(x) increasing or decreasing?Sign of the slope of the tangent linex = 0.5increasingpositivex = 1increasingpositivex = 3increasingpositivex = 10increasingpositivex = 15increasingpositivex = 20increasingpositivex = 25increasingpositivex = 30increasingpositive Repeat the process for drawing a tangent at a point for the function.  QUOTE   Use the window settings [-3, 6] by [-2, 2]. Enter the results in the table below. Remember to drag the point as close to the x-value as you can. Coordinate PointIs y = sin(x) increasing or decreasing?Sign of the slope of the tangent linex = -2.7decreasingnegativex = -2.5decreasingnegativex = -1increasingpositivex = -0.5increasingpositivex = 0.5increasingpositivex = 1increasingpositivex = 2.5decreasingnegativex = 3decreasingnegativex = 4decreasingnegativex = 5increasingpositive (a) Looking at all of the charts, what conclusion can be made regarding the sign of the slope of the tangent when the function is increasing? The slope is positive. (b) When the function is decreasing? The slope is negative. Where do you think that the sign of the tangent switches from positive to negative? When the slope is neither positive or negative. When the slope equals zero. Without graphing, can you tell if a function is increasing or decreasing by looking a given point on the function? Consider y = 2x2 + 4x – 3. Take the derivative by hand at the points x = -2 and at x = 3. Is it increasing or decreasing at each of those points? Show all of your work.   QUOTE   The slope is negative, consequently the curve is decreasing at  QUOTE    QUOTE   The slope is positive, consequently the curve is increasing at  QUOTE   Part Two: Slopes of Tangents and Local Extrema NOTE: Local extrema are the mimimums or maximums of a function on a given interval. Graph  QUOTE  , on a new graph screen using a window setting of [-10, 10] by [-10, 10]. Draw a point on the curve and draw a tangent to that point. Drag the tangent to the points where the graphs reach a minimum value. Fill in the chart below: Refer to steps 1-5 from Part One. Graph the point and drag it to the minimum value. You know you are at the minimum because a lower case m in a box appears to the lower right of the “hand.” Remember that if you cannot read some of the information because it is all bunched together, you can drag the coordinates and slopes to another area of the screen. Refer to step 2 in part one. Coordinates of MinimumValue of the slope at the minimumThe line is increasing, decreasing, horizontal, or vertical…(0, 0)0horizontal Graph  QUOTE   on a new graph screen using, the same window settings. Draw a point on the curve and draw a tangent to that point. Drag the tangent to the points where the graphs reach a maximum value. Fill in the chart below: Graph the point and drag it to the maximum value. You know you are at the maximum because an upper case M in a box appears to the lower right of the “hand.” Coordinates of MaximumValue of the slope at the maximumThe line is increasing, decreasing, horizontal, or vertical…(0, 0)0horizontal Graph  QUOTE   on a new graph screen using the window [-3, 6] by [-2, 2]. Draw a point on the curve and draw a tangent to that point. Drag the tangents to the points where the graphs reach a maximum or minimum value. Fill in the chart below: Min or MaxCoordinates of Min or MaxValue of the slope at the min or maxThe line is increasing, decreasing, horizontal, or verticalMinimum(3Ą/2, -1)0horizontalMaximum(Ą/2, 1)0horizontalMinimum(-Ą/2, -1)0horizontal Now make a conclusion about your observations from #1-3. Fill in the blanks: At an extrema, the slope of the tangent is __ 0 __ and the tangent line is __ horizontal ________. Now we want to examine the behavior of a function before and after an extrema point. Consider  QUOTE  ,. From part two, #1, the coordinates of the minimum were (0, 0). Fill in the table below, using information from part one and part two, as well as your graph: intervalf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-", 0)Decreasingnegative(0, ")Increasingpositive Conclusion: The point (0, 0) is a minimum or maximum? The point is a minimum. Consider  QUOTE  ,. From part two, #2, the coordinates of the maximum were (0, 0). Fill in the table below, using information from part one and part two, as well as your graph: intervalf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-", 0)Increasingpositive(0, ")Decreasingnegative Conclusion: The point (0, 0) is a minimum or maximum? The point is a maximum. The function  QUOTE  on the given interval had two minimums and one maximum. From part 2 #3, what were the coordinates of these three points? Mimimums: Maximum: Fill in the table below, using information from part one and part two, as well as your graph: intervalsf(x) increasing or decreasing?Is the slope of tangent positive or negative?(-3, -1.5708)decreasingnegative(-1.5708, 1.5708)increasingpositive(1.5708, 4.7123)decreasingnegative(4.7123, 6)increasingpositive Make a conclusion about determining if a point is a maximum or minimum based on your observations: A point is a maximum if the slope at the point is ___ 0 _______ and the slope of the tangents switch from __ positive __ before the point to ___ negative ___after the point. A point is a minimum if the slope at the point is ____0____ and the slope of the tangents switch from __ negative ____before the point to ___ positive ___after the point. Consider the other functions from part 1: y = x3, y = ex, and  QUOTE  . Look at the graphs and your tables from part one. Do these functions have any extrema? If so, are they minimums or maximums? They have no extrema. Explain your answer based on your conclusions from part two, #8. The curves are always increasing. The slopes of the tangent lines never switch signs. How can find extrema without looking at the graph? Consider y = 2x2 + 4x – 3. Take the derivative by hand. Based on your answers from #8, find where an extrema occurs. An extrema can occur when the slope of the tangent line equals zero. Determine if the extrema is a maximum or minimum. Show all work and explain your answer.  The derivative (slope of the tangent line = 0 at x = -1).  QUOTE   The slope is negative, consequently the curve is decreasing at  QUOTE    QUOTE   The slope is positive, consequently the curve is increasing at  QUOTE   Thus, the function has a minimum at x = -1, since the slopes change from negative to positive. Part Three: Tangents Above or Below the Curve and Concavity Tangents can lie above or below the curve. All tangent lines that lie below the curve have y-values that are less than the corresponding y-values on the function. In other words, if f(x) are the y-values on the curve and T(x) are the y-values on the tangent curve, then T(x) d" f(x) for all values of x. All tangents lines that lie above the curve have y-values that are more than the corresponding y-values on the functions. In other words, T(x) e" f(x) for all values of x. Let s revisit the function  QUOTE  ,. As you move the point along the graph, are the tangents above the graph or below the graph? They are below the graph. What is the general shape of the function, specifically looking at whether it opens upward or downward? The graph opens upward. A graph that opens upward is referred to as being concave up. A graph that opens downward is concave down. Is  QUOTE   concave up or concave down?  QUOTE   is concave up. Now look at  QUOTE  . As you move the point along the graph, are the tangents above the graph or below the graph? They are above the graph. Is  QUOTE   concave up or down?  QUOTE   is concave down Now look at y = ex. As you move the point along the graph, are the tangents above the graph or below the graph? They are below the graph. Is y = ex concave up or down? y = ex is concave up Now look at  QUOTE  . As you move the point along the graph, are the tangents above the graph or below the graph? They are below the graph. Is  QUOTE   concave up or down?  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With respect to tangent lines, a function if concave down when: the tangent lines lie above the graph. Part Four: Points of Inflection and Concavity Some functions won’t have a single concavity. Let’s look at some examples of those functions. As you move the point along the graph, the tangents lines may switch positions. The point at which the concavity switches is called the point of inflection. At this point, some of the points of the tangent line are above the curve and some of the points are below. Consider y = x3. Drag the tangents along the curve to find at which point on y = x3 this occur? This occurs at x = 0. On what interval is y = x3 concave up? On what interval is y = x3 concave down? It is concave down on (-", 0) and concave up on (0, ") Now, consider  QUOTE   on [-3, 6] by [-2, 2]. On what interval(s) is  QUOTE   concave up and on what interval(s) is  QUOTE   concave down? The curve is concave up on the intervals (-3, 0) and (Ą, 6) The curve is concave down on the interval (0, Ą) What are the points of inflection and where do they occur? The points of inflection are where the curve changes concavity and occur at x  QUOTE   . Extension One: Position, Acceleration, and Velocity The velocity of a particle is defined to be the rate of change in position with respect to time. Specifically, instantaneous velocity at any time is given by the equation  QUOTE   The acceleration of a particle is defined to be the rate of change in velocity with respect to time. Specifically, instantaneous acceleration at any time is given by the equation  QUOTE   A particle is said to be speeding up when its velocity and acceleration are in the same direction (both positive or both negative). A particle is said to be slowing down when its velocity and acceleration are in opposite directions (either positive or negative). A particle moving in the y-direction has a position given at any time t by the function:  QUOTE  . (Suggested viewing window for this problem is [-1, 12] x [-40, 50]) During the first 8 seconds of the particle’s motion: When is the object falling? On the interval (1.6906, 6.3094) When is the object rising? On the intervals (0, 1.6906) and (6.3095, 8) When does the object reach its maximum height? If students consider a local maximum, then x = 1.6906, the height is 16.3168. In addition to the relative extrema, students should consider the endpoints. When does the object reach its minimum height? When x = 6.3094, height is -8.31681. Again, in addition to the relative extrema, students should consider the endpoints. When is its velocity increasing? The velocity is increasing on the interval (4, 8). This should be the point of inflection and can be an estimate close to x = 4. When is its velocity decreasing? The velocity is decreasing on the interval (0, 4). This should be the point of inflection and can be an estimate close to x = 4. When does the object reach maximum velocity? The maximum velocity is x = 0. This solution is found by considering the endpoints of x = 0 and x = 8. Since from parts e and f from above, x = 4 is a minimum velocity. Since the velocity at x = 0 is more than the velocity at x = 8, then x = 0 yields a maximum velocity. Extension Two: Linear Approximation and Tangent Lines Graph the equation y = x3 + 2x2 - 3x – 4 on the window [-10, 10] by [-10, -10]. Draw a point on the curve, draw a tangent line through that point, and find the equation of the tangent line, using the TI-Nspire. To draw a point on the curve and draw a tangent through that point, refer to Activity 1. For the equation of the tangent, choose the menu icon, select 1:Action’!6:Coordinates and Equations Move the cursor to the point and select enter and the line will be highlighted and the equation appears lightly. Hit enter a second time and the equation will darken. Hit the esc key. (You can drag this equation to an empty area of the graph by following the directions in Activity 1) We know the point (1, 2) is on the curve. The equation to the tangent line at the point x=1 is given by  QUOTE   We can use this tangent line to approximate values of the function close to the point x=1. For example, the estimate,  QUOTE   Notice that actual value,  QUOTE   Also notice that the estimate,  QUOTE  , whereas the actual value,  QUOTE  . In both cases, the estimates are less than the actual values. Looking at the tangent line at x = 1, we could have predicted this underestimation. Why? The tangent line is below the curve, so any point on the tangent line will be less than the true point on the curve. Now consider the point (-2,2) on the curve. The equation of the tangent line at this point is  QUOTE   What is the linear approximation of  QUOTE   at the point x = -2.08? The linear approximation of x = -2.08 is y = 1.92 What is the actual value at this point? ^Š`ŠdŠ€Š‚ŠŽŠ ‹ ‹(‹T‹ˆ‹Š‹Ų‹Ś‹ś‹hŒ†ŒˆŒ–Œ˜ŒšŒœŒžŒ Œ¢ŒńåÖĘåÖĘÖ³å¦å¦å³å‘€s€‘f‘XåE%jhEE@hEE@CJOJQJUaJh»ļCJOJPJQJaJjhEE@hEE@EHś’Uj|ü hEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJh»ļCJH*OJQJaJ%hh»ļB* CJOJQJaJphĄhPG]h»ļCJH*OJQJaJhPG]h»ļCJOJQJaJh»ļCJOJQJaJh1jžh»ļCJOJQJaJ%‰bŠdŠ(‹T‹V‹ų‹ś‹hŒjŒ¦؍ Ž‚Ž„ŽśŽüŽ˜øśńéįśŲśįśéśŠŠśńńÄļ·gd¶$a$gd¶ $„Š^„Ša$gd»ļ$a$gd»ļ„Š`„Šgd»ļ$a$gd»ļ & Fgd¶„Š^„Šgd»ļgd»ļjlz|~€‚„ˆ؍‚ŽųŽśŽüŽ˜ńäńŃÄŃø£’…’£x£jøWøHøWhżh»ļCJOJQJaJ%hĻW3h»ļB* CJOJQJaJphĄh»ļCJOJPJQJaJj’hEE@hEE@EHś’UjōfhEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJh»ļCJOJQJaJjVLhEE@hEE@EHś’U%jhEE@hEE@CJOJQJUaJjø1hEE@hEE@EHś’UhEE@hEE@CJOJQJaJ˜š؏Ŗ¬®°²¶ø½¾ŐʐĒȐęŃÄŃę·ę¢ƒyhSB5BSj„ąhEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJ h;kšh¶CJOJPJQJaJh˜^ōh¶6PJh¶5CJOJQJaJ#h»ļB* CJOJPJQJaJphĄ)hĻW3h»ļB* CJOJPJQJaJphĄjZ¾hEE@hEE@EHņ’Uj0œhEE@hEE@EHņ’U)hEE@hEE@B* CJOJPJQJaJphĄ2jhEE@hEE@B* CJOJPJQJUaJphĄȐɐŹ~‘‘†‘‡‘ˆ‘‰‘Š‘‹‘Œ‘’’”’Ģ’ģ’ķ’ī’õ’ö’÷’ų’ł’ś’ū’ü’“ “ņŻĢŻ»®»Ż”Ż“Ģ“„x„Ż»k»Ż^ŻĢ“TGh¶h¶CJPJaJh¶CJPJaJj|€hEE@hEE@EHņ’Ujd]hEE@hEE@EHņ’Uh¶CJOJQJaJh;kšh¶CJOJQJaJh¶CJOJPJQJaJj+>hEE@hEE@EHś’UjņhEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ h;kšh¶CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJj»’hEE@hEE@EHś’UĖŒ‘“’”’ķ’ü’F“{“—“ø“ӓ”/”Ė”ś”t•••–9–öööńéŁŃČĄ»³»³»³»³»« & Fgd¬'ˆ & Fgd¶gd¶ & Fgd¶„Š^„Šgd¶dhgd¶„ „Šd^„ `„Šgd¶ & Fgd¶gd¶„Š`„Šgd¶ ““F“X“y“z“—“·“ø“Ņ“ӓߓž“’“”.”/”{”Ź”Ė”ł”ś”’”s•t•”•••––öéŻĪĀĪ­˜‡y­d­˜‡y­d˜‡y­d˜‡yO˜)h Ņh¬'ˆB* CJOJPJQJaJphĄ)h Ņh…qęB* CJOJPJQJaJphĄh¶CJOJPJQJaJ hÉ=œh¶CJOJPJQJaJ)h Ņh¶B* CJOJPJQJaJphĄ)h ŅhQ(uB* CJOJPJQJaJphĄh¶CJOJQJaJhä.ah¶CJOJQJaJh…qęCJOJQJaJh¶h¶CJPJaJh…qęCJPJaJ––8–9–¼–č–é–ū—ü—ž—3˜4˜:˜M˜ļŽÉ“£•‚reUD8)hį šhä.aCJOJQJaJhä.aCJOJQJaJ hžÅhä.aCJOJPJQJaJhžÅhä.a5CJOJQJaJhä.a5CJOJQJaJh¬'ˆB*CJOJQJaJph’%hh uh¬'ˆB* CJOJQJaJphĄh¶CJOJPJQJaJ hÉ=œh¶CJOJPJQJaJ)h Ņh¬'ˆB* CJOJPJQJaJphĄ)h¬'ˆh¬'ˆB*CJOJPJQJaJph’ h¬'ˆh¶CJOJPJQJaJ h¬'ˆh¬'ˆCJOJPJQJaJ 9–¼–é–A—4˜™a™Jš,›˜›ŗ›BœjĀžZžŠž>ŸˆŸśņśśźįįįįįÕĶÄÄÄ»¶Ķ­„h`„hgdä.agdä.a„Š`„Šgdä.a„Š^„Šgdä.a & Fgd¶ „>ždš]„>žgdä.a„>ž]„>žgdä.a & Fgd¶ & Fgd¶gd¬'ˆM˜N˜S˜T˜’˜™™’™šø›ŗ›BœWœXœŖœ«œ²œ³œ“œµœ¶œ·œ¹œļąļąŌąĢĮĢµĢ¦š¦…tgt…Z…Lhä.aCJOJPJQJaJjI¼hEE@hEE@EHś’Uj”£hEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJhä.aCJOJQJaJh;kšhä.aCJOJQJaJh¢?hä.aCJH*aJh¢?hä.aCJaJhä.aCJaJhTd§CJOJQJaJhį šhä.aCJOJQJaJhį šhä.aCJH*OJQJaJ¹œ0189:;<=BQZ[bcdefgŠ‹’“”•–—«³“»ļįļĢ»®»Ģ”Ģļ“ļĢ»†»ĢyĢļĢ»l»Ģ_Ģ“ļĢ»jz]hEE@hEE@EHś’UjŽAhEE@hEE@EHś’UjG'hEE@hEE@EHś’Uj° hEE@hEE@EHś’Uhh uCJOJPJQJaJjךhEE@hEE@EHś’UjžŌhEE@hEE@EHś’U hEE@hEE@CJOJPJQJaJ)jhEE@hEE@CJOJPJQJUaJhä.aCJOJPJQJaJ h;kšhä.aCJOJPJQJaJ »¼½¾æĄĮĀ֝ųZžĻžŠžģž.Ÿ/Ÿ0Ÿ1Ÿ8Ÿ9Ÿ:Ÿ;Ÿ<Ÿ=ŸbŸcŸjŸņįĢæĢ® ®’®€k]® ®ĢįPįĢCĢ®ĢįjŖŹhEE@hEE@EHś’Uj€±hEE@hEE@EHś’Uhj„CJOJPJQJaJ)hh uhä.aB* CJOJPJQJaJphĄ#hh uB* CJOJPJQJaJphĄhh uCJOJPJQJaJhä.aCJOJPJQJaJ h;kšhä.aCJOJPJQJaJjK•hEE@hEE@EHś’U)jhEE@hEE@CJOJPJQJUaJ hEE@hEE@CJOJPJQJaJjyhEE@hEE@EHś’UjŸkŸlŸmŸnŸoŸˆŸµŸ¹ŸŗŸʟĶŸāŸ¼1¼2¼3¼S¼€¼ņįĢæĢ®œŠu®g®eS>®0®h CJOJPJQJaJ)h h B* CJOJPJQJaJphĄ#h B* CJOJPJQJaJphĄUhh uCJOJPJQJaJ)hh uhä.aB* CJOJPJQJaJphĄ#h_1B* CJOJPJQJaJphĄ#hj„B* CJOJPJQJaJphĄ h;kšhä.aCJOJPJQJaJjĢłhEE@hEE@EHś’U)jhEE@hEE@CJOJPJQJUaJ hEE@hEE@CJOJPJQJaJjŌćhEE@hEE@EHś’UˆŸŗŸāŸ2¼Ž¼Ļ¼ü¼½{½ī½P¾dæöķöåöķķķÜåÓ„Š^„Šgd_1„h^„hgd_1 & Fgd¶„h`„hgdä.a„h`„hgd_1 The actual value of y at x = -2.08 is y = 1.89389 Similarly for the point x = -1.997: what is the linear approximation and the actual value? The linear approximation is 2.003. The actual value is 2.00296. Are the two linear estimates over or under? Both linear estimates are over. What do we notice about the relationship between the tangent line and the curve at this point? The tangent line is above the curve, so the points on the tangent line would be more than the points on the curve. What can you conclude about linear approximations using tangents and the concavity of the curve? If the curve is concave up, the linear approximations would be less than the actual value because the tangent lines are below the curve. 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